But MathFest also got me thinking about my upcoming job search, perhaps because one of my friends let me know his department will be hiring in Algebra this year. And I started to wonder how much a tenure-track job in mathematics, or in any academic field, for that matter, is actually worth. I mean, could you determine an actual dollar value for the worth of a a tenure-track academic position?

Perhaps I’ve been listening to too much Planet Money, Marketplace, and other economics reporting on public radio, or perhaps I’m taking the name ‘job market’ too seriously, but it seems to me that the value of a tenure-track job is whatever an individual would be willing to pay today in order to secure such a position. If you don’t think that fresh and recent PhDs wouldn’t fork over a chunk of change for the opportunity to secure a tenure-track job, then you haven’t been through the academic job application process. (I was going to append the word ‘lately’ to the end of that sentence, but then decided my point was valid without it.)

DISCLAIMER: I am not offering any up-front cash payments in return for a tenure-track job. I just think this is an interesting question.

Now, if I were going to sit down and try to figure out the value of a particular tenure-track position, I suppose I might take a cue from the second semester Applied Calculus course I taught at UVA. I could think of the job as an income stream, and then try to compute the present value of that income stream. Call the present value PV. Perhaps this is the value of the tenure-track position? If you pay the potential employer PV dollars, then they’ll break even over the course of your employment. Financially, it’s a win-win for them. On the other hand, PV is probably quite large, more than a fresh or recent PhD might have on hand. But perhaps a bank would make you a loan so you could secure the tenure-track position. Whatever you ended up paying for the job, you’d want to make sure that the difference between your income and the debt you incur to get the job is still great enough that you have enough money to buy groceries, pay rent, and generally make a living. So perhaps no one in their right mind would pay as much as PV.

A school has 1000 lockers, numbered consecutively from 1 to 1000. The school also has exactly 1000 students. The first student walks into the school and opens every locker door. Then the second student closes the door on every locker labeled by an even number. Then the third student switches the doors on every locker labeled by a multiple of three (that is, opens the door if it is closed, and closes the door if it is open). Then the fourth student switches the doors on every locker labeled by a multiple of four. Then the fifth student switches the doors on every locker labeled by a multiple of five. And so on, for all the rest of the students. After all 1000 students have walked through the school, which locker doors will remain open?

Today I did a Google search for Inception, and discovered by chance that it is playing this weekend at the theater in the student union on campus. Admission is $1 for students, $2 for everyone else. I made an attempt to go to tonight’s 9pm show, but the line was very long, and there was only one person working the ticket window. I stood in line until 9:15 before deciding that even if they started the film late, I probably wasn’t going to get a ticket and sit down before the show started. Perhaps I’ll try again for the 3pm show tomorrow afternoon.

I don’t remember the student theater at UVA ever being that crowded, but maybe they charged more than $1 for student admission.

We argue by mathematical induction. Clearly, all horses in any set of 1 horses are all the same color. Now assume that the result is true for any collection of n horses, and consider a collection of n+1 horses. In the collection of n+1 horses, we know that horses 1, 2, …, n are all the same color (by the induction hypothesis), as are horses 2, 3, …, n+1 (since this is another collection of n horses). These two subcollections have horses 2, 3, …, n in common. Thus, all horses in the collection 1, 2, …, n, n+1 are the same color. Therefore, by the principle of mathematical induction, all horses are the same color.

Out of the 29 students, only two correctly (or very nearly correctly) pointed out the flaw in this logic. Can you? It’s easy to say that not all horses are the same color. It’s more difficult to say exactly how the principle of mathematical induction was misapplied.

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1. On Monday, I tell the students that there will be an in-class quiz this week, and that if I don’t give a quiz on Wendesday, then I will give a quiz on Friday.
2. On Wednesday, I give the students a quiz.
3. On Friday, I give the students a second quiz. The second quiz consists of the following question:

Let P be the statement “There is no quiz on Wednesday,” and let Q be the statement “There is a quiz on Friday.” On Monday, your instructor made the statement . Based on his actions this week, and based on the fact that you are now taking a quiz on Friday, what is the truth value of the statement ?

1. Possibly get a new TV. Upgrade from the 12″ model to something larger.
2. Prepare a week’s worth of lectures discussing elementary aspects of mathematical logic (statements, negations, truth tables, implications, contrapositives, tautologies, contradictions, equivalences).
3. Edit PDF proof of my recently accepted paper.
4. Read about compact Lie groups. Finally open and read the textbook from the course on Lie groups that I took back in Fall 2007.
5. Think about the relation between the cohomology of compact Lie groups and the cohomology of simple Lie algebras. Post a related question on MathOverflow and hope someone smarter than me knows the answer.
6. Make revisions to my even-more-recently accepted paper.
7. Play Metroid on the Nintendo emulator.

I’m writing from the Project NExT meeting in Pittsburgh, PA. Project NExT is a professional development program of the Mathematical Association of America. The NExT in Project NExT stands for New Experiences in Teaching. I can’t say exactly what those new experiences are yet, but I can make a few observations after my first day of the meeting.

First, of all the participants here, postdocs like myself are in the minority. Most of the NExTers hold tenure track positions. I envy the people who don’t have a predetermined definite end date to their employment. One way or another, my current job will be over by Spring 2012.

Second, it seems to me that most of the NExTers hold positions at four year colleges, and especially at small liberal arts schools. That is the kind of position I’m trying to set myself up for, so perhaps I’m ahead of the game by going through Project NExT while still a postdoc, rather than waiting until I have a tenure track job. Actually, I’m hoping that my Project NExT experiences will be a boon as I go on the job market for a tenure track job (in addition to being a boon as I finish my postdoc).

Finally, I’ve already met several people who have experience teaching courses for prospective elementary and middle grades teachers. That’s exactly what I anticipate doing next Spring. They described successful experiences with such courses, so that gives me more confidence that I could do a good job, too.

More reports from this workshop and from MathFest 2010 may be forthcoming. Stay tuned. Or don’t. It’s up to you.

Note added in proof: Evidently Kleshchev himself made the elliptical orbit joke. I didn’t hear him make it, but only heard Jon repeat the joke a moment later.

Let be a finite group, and let be a field. The group ring is the associative, unital -algebra with vector space basis indexed by the elements , and with multiplication induced by the multiplication in . The group ring is a also a Hopf algebra, with comultiplication defined on the basis vector by , and with counit defined on the basis element by .

Now consider the linear dual . Since is a Hopf algebra, is an algebra. Given and , the product is the linear map defined on the basis element by .

Given , write for the linear function which maps , and which maps all other elements of to zero. The elements for form a basis for . Now observe that if , then . Also, if and , then .

The identity element in the ring is the sum . From the remarks in the previous paragraph, we see that this is a decomposition of into orthogonal idempotents. Now is a decomposition of the ring into a direct product of simple rings. This implies that every -module is semisimple, i.e., a direct sum of simple modules. Furthermore, since the terms in the direct product decomposition of are all just isomorphic to , those simple modules are nothing more than -vector spaces!

The representation theory of the group ring can be very complicated (particularly if is a field of positive characteristic). But the representation theory of is exceedingly simple! In particular, the ring doesn’t have any cohomology, so to speak. Given left -modules and , all of the higher -groups () vanish!